REASONING AND SOLUTION

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1 39. REASONING AND SOLUTION The heat released by the blood is given by Q cm T, in which the specific heat capacity c of the blood (water) is given in Table Then Therefore, T Q cm 2000 J 0.8 C [4186 J/(kg C )](0.6 kg) T f T i T 36.2 C

2 40. REASONING Since the container of the glass and the liquid is being ignored and since we are assuming negligible heat exchange with the environment, the principle of conservation of energy applies. In reaching equilibrium the cooler liquid gains heat, and the hotter glass loses heat. We will apply this principle by equating the heat gained to the heat lost. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount T is given by Equation 12.4 as Q cm T, where c is the specific heat capacity. In using this equation as we apply the energy-conservation principle, we must remember to express the change in temperature T as the higher minus the lower temperature. SOLUTION Applying the energy-conservation principle and using Equation 12.4 give c m T c m T Liquid Liquid Glass Glass Heat gained by liquid Heat lost by glass Since it is the same for both the glass and the liquid, the mass m can be eliminated algebraically from this equation. Solving for c Liquid and taking the specific heat capacity for glass from Table 12.2, we find c Liquid Liquid ( ) ( ) ( 53.0 C 43.0 C ) c 840 J/ kg C 83.0 C 53.0 C Glass T Glass 2500 J/ kg C T ( )

3 41. SSM REASONING According to Equation 12.4, the heat required to warm the pool can be calculated from Q cm T. The specific heat capacity c of water is given in Table In order to use Equation 12.4, we must first determine the mass of the water in the pool. Equation 11.1 indicates that the mass can be calculated from m ρv,where ρ is the density of water and V is the volume of water in the pool. SOLUTION Combining these two expressions, we have Q cρv T,or Q [ 4186 J/(kg C ) ]( kg/m 3 )(12.0 m 9.00 m 1.5 m) (27 C 15 C) J Using the fact that 1kWh J, the cost of using electrical energy to heat the water in the pool at a cost of $0.10 per kwh is ( $0.10 J) J $230.

4 44. REASONING AND SOLUTION We require that the heat gained by the cold water equals the heat lost by the hot water, i.e., Therefore, Q cw Q hw c cw m cw T cw c hw m hw T hw Since the specific heat capacity is that of water in each case, c cw c hw,then Then m cw (36.0 C 13.0 C) m hw (49.0 C 36.0 C) m cw (0.57)m hw We also know that m cw +m hw 191 kg. Substituting for m cw, and solving for m hw,we have 191 kg m hw 121 kg 1.57

5 53. REASONING AND SOLUTION We want the same amount of heat removed from the water as from the ethyl alcohol, i.e., or Q water Q alcohol (ml f ) water (ml f ) alcohol Then, taking values of the latent heats of fusion for water and ethyl alcohol from Table 12.3, we find m alcohol m water ( L f ) water ( ) alcohol 3.0 kg L f ( ) J/kg kg J/kg

6 54. REASONING As the body perspires, heat Q must be added to change the water from the liquid to the gaseous state. The amount of heat depends on the mass m of the water and the latent heat of vaporization L v, according to Q ml v (Equation 12.5). SOLUTION The mass of water lost to perspiration is 4186 J ( 240 Calories) Q 1Calorie m L J/kg v 0.42 kg

7 58. REASONING Since the container is being ignored and since we are assuming negligible heat exchange with the environment, the principle of conservation of energy applies in the following form: heat gained equals heat lost. In reaching equilibrium the colder aluminum gains heat in warming to 0.0 ºC, and the warmer water loses heat in cooling to 0.0 ºC. In either case, the heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount T is given by Equation 12.4 as Q cm T, wherec is the specific heat capacity. In using this equation as we apply the energy-conservation principle, we must remember to express the change in temperature T as the higher minus the lower temperature. The water that freezes into ice also loses heat. The heat Q lost when a mass m of water freezes is given by Equation 12.5 as Q ml f,wherel f is the latent heat of fusion. By including this amount of lost heat in the energy-conservation equation, we will be able to calculate the mass of water that is frozen. SOLUTION Using the energy-conservation principle and Equations 12.4 and 12.5 gives c m T c m T + m L Aluminum Aluminum Aluminum Water Water Water Ice f, Water Heat gained by aluminum Heat lost by water Heat lost by water that freezes Solving for m Ice, taking values for the specific heat capacities from Table 12.2, and taking the latent heat for water from Table 12.3, we find that m Ice c m T c m T L Aluminum Aluminum Aluminum Water Water Water f, Water ( ) ( ) ( ) J/ kg C kg 0.0 C 155 C J/kg ( ) ( )( ) 4186J/ kg C 1.5 kg 3.0 C 0.0 C J/kg kg

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